# Math Help - Deriving Mollweide's Formula

1. ## Deriving Mollweide's Formula

For this post, A is alpha, B is beta, and Y is gamma, and represent the angles of the triangle. Little a, b, and c represent the sides opposite them, respectively. The form of the formula is:
Code:
a - b         sin(1/2[A - B]
-----   =    -------------
c            cos([1/2]Y)

There is a step in the derivation that I don't understand. Here is the derivation from my solutions book, and so this post doesn't get too protracted, I'll stop at the point that I don't understand:

a - b        a    b
-----  =     -- - --
c          c    c

=  sin A     sin B
----   -  ----
sin Y     sin Y

= sin A - sin B
-------------
sin Y

= 2sin( (A-B) / 2) cos( (A+B) / 2)
------------------------------    makes use of sum to product formula
sin (2Y/2)

= 2sin( (A-B) / 2) cos( (A+B) / 2)
--------------------------------
2sin(Y/2)cos(Y/2)

= sin( (A-B) / 2) cos(PI/2 - Y/2)
--------------------------------
sin(Y/2)cos(Y/2)

How does cos((A+B)/2) get turned into cos(PI/2 - Y/2)?

2. Well you do know that $\alpha + \beta + \gamma = \pi \quad \Rightarrow \quad \alpha + \beta = \pi - \gamma.$

Now divide by 2.

3. I didn't see that for some reason. Thanks.